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Let $A, B \in L(V)$ (or equivalently $n \times n$ matrix), where $V$ is a $n$-dimensional vector space. There are a multiple of proofs why $\det(A)\det(B) = \det(AB)$, but I couldn't find a satisfactory that is intuition-appealing.

First, the most imminent motivation for determinant is volume: $\det(A)$ is oriented volume of parallelepiped consisting $n$-column vectors of $A$. Given three matrices $A, B, AB$, we have three distinct parallelepiped each of which has oriented volume $\det(A)$, $\det(B)$, and $\det(AB)$, respectively.

Second, the given $A, B \in L(V)$ as in linear transformation, $AB$ is simply a composition of linear transformation.

I am trying to relate these two concepts, but it feels that the gap between two concepts are too broad. Is there a nice interpretation that fills the gap between these two concepts ?

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  • $\begingroup$ A slightly different question whose answers may answer your question too: math.stackexchange.com/questions/3352492/… $\endgroup$
    – Hans Lundmark
    Jun 11, 2021 at 7:20
  • $\begingroup$ See this page : mathinsight.org/determinant_geometric_properties $\endgroup$
    – Mathematician 42
    Jun 11, 2021 at 7:27
  • $\begingroup$ $\det(A)$ is not a volume. You may treat it as the volume of the parallelepiped whose vertices are the origin and the endpoints of the column vectors of $A$ if you assume that the "unit (hyper)cube" in the current basis has volume $1$, but the determinant is not a volume by nature, because it is dimensionless (i.e. it doesn't carry any measurement unit). As stated in an answer below, it is the relative change of volume (it's a ratio, hence unitless) when an object undergoes the linear transformation represented by $A$. $\endgroup$
    – user1551
    Jun 11, 2021 at 9:49
  • $\begingroup$ the second answer in this link this the proof i prefer for this result. it assumes some basic knowledge of multi linear forms to understand, but i think it gives the most convincing answer. $\endgroup$
    – C Squared
    Jun 11, 2021 at 11:33

1 Answer 1

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Here's a slightly modified interpretation of the $\det(A)$ as a volume. $\det(A)$ is more generally the ratio between volumes in the input and output space.

Indeed, in linear transformations, volumes scale uniformly, and the scale factor from domain to codomain of any set transformed by a linear map (not just the fundamental parallelepiped) gives the same ratio, which we know is the determinant.

Thus, take an arbitrary set in the vector space. The determinant of the composition mapping $AB$ is the ratio of the volumes of the given set under the transformation. $B$ first scales the set's volume by $\det(B)$, and then $A$ does by $\det(A)$, so the total scale factor, which is $\det(AB)$ is just the product of these intermediate scale factors. Hence $$\det(AB) = \det(A) \det(B)$$

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  • $\begingroup$ What is an "old" set ? $\endgroup$
    – Jean Marie
    Jun 11, 2021 at 9:22
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    $\begingroup$ The phrase "any old set" is an expression meaning "an arbitrary set". I'll clarify. $\endgroup$
    – Cade Reinberger
    Jun 11, 2021 at 9:23
  • $\begingroup$ Thanks. I had never seen this 'coloquial' expression. $\endgroup$
    – Jean Marie
    Jun 11, 2021 at 9:32

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